If there are no degeneracies, do the eigenvectors of an invertable matrix have to be orthoganal? If we have a matrix A with some eigenvectors $ev_1, ev_2, ...$
and eigenvectors are independent but not orthogonal, we can represent 
$k(ev_1)+v=ev_2$
where v is a vector orthoganal to $ev_1$, and k is a constant. 
then we can do
$$A(k(ev_1)+v)=A(ev_2)$$
$$
=\lambda_1(k)(ev_1)+Av =\lambda_2(ev_2)=\lambda_2(k(ev_1)+v)=\lambda_2(k(ev_1))+\lambda_2(v)$$
thus
$$\lambda_1(k)(ev_1)+Av=\lambda_2(k(ev_1))+\lambda_2(v)$$
reorder refactor
$$( \lambda_1 -\lambda_2)(k)ev_1=\lambda_2v-Av$$
since LHS is a  $\lambda_1$ ev, then RHS is also a $\lambda_1$ ev. If v is an ev then there is a clear degeneracy because it lies on the same plane as ev_1 and ev_2. 
If vs is not an ev, then that implies that A cannot take v off the plane of v and ev_1 (and ev_2 by that regard. ) Does this imply degeneracy? 
 A: No. 
Example: 
\begin{bmatrix}
1 & 1\\ 
0 & 2
\end{bmatrix}
has eigenvector $(1, 1)$ with eigenvalue $2$, and $(1, 0)$ with eigenvalue $1$.
As Doug M notes, if the matrix is symmetric, then the eignevectors for distinct eigenvalues are orthogonal. (For the same eigenvalue, they need not be orthogonal; for example, the $2 \times 2$ identity has all nonzero vectors as eigenvectors for $1$.) 
A: Consider $A=\begin{bmatrix} 1&1\\ 0&2 \end{bmatrix}$. Clearly $Ae_{1}=e_{1}$ and $A(e_{1}+e_{2})=2(e_{1}+e_{2}).$ We can see that here $v=e_{2},$ and $(2I-A)v=\begin{bmatrix} 1&-1\\ 0&0\end{bmatrix}v=-e_{1}=(1-2)e_{1},$ as you noted above, and indeed, $A$ does not move $v$ out of the $e_{1}/e_{2}$-plane (or the $e_{1}/(e_{1}+e_{2})$-plane), but $A$ is clearly not degenerate ($Ax=0$ implies $2x_{2}=0,$ so $x_{2}=0,$ and $x_{1}+x_{2}=0,$ so $x_{1}=0$).
It's worth pointing out that since matrix multiplication is linear, any hyperplane will be mapped to a hyperplane: $A(\alpha_{1} x_{1}+\alpha_{2}x_{2}+\cdots+\alpha_{n}x_{n})=\alpha_{1}Ax_{1}+\cdots+\alpha_{n}Ax_{n},$ so $A$ takes $\mathrm{span}(\{x_{1},\ldots,x_{n}\})$ to $\mathrm{span}(\{Ax_{1},\ldots,Ax_{n}\}).$ A matrix is degenerate when it maps some hyperplanes to hyperplanes of strictly lower dimension. Indeed, if we consider $A=\begin{bmatrix} 1&1\\ 0&0\end{bmatrix},$ then $Ae_{1}=e_{1},$ but $A(e_{1}-e_{2})=0,$ so $A$ maps the $e_{1}/e_{2}$-plane to $\mathrm{span}(\{e_{1}\}),$ which has strictly lower dimension.
